Step by step solution to "Evaluate the integrals ∫0π4sec2x(1+7tan⁡x)23"

beljuA

Answered question

2021-10-12

Evaluate the integrals

\int_{0}^{\frac{π}{4}} \frac{\sec^{2} x}{{(1 + 7 \tan x)}^{\frac{2}{3}}}

aprovard

Skilled2021-10-13Added 94 answers

To evaluate:

\int_{0}^{\frac{π}{4}} \frac{\sec^{2} x}{{(1 + 7 \tan x)}^{\frac{2}{3}}} d x

Let substitute

t = (1 + 7 \tan x)

\frac{d t}{d x} = \frac{d}{d x} (1 + 7 \tan x)

\frac{d t}{d x} = 7 \sec^{2} x d x

\frac{d t}{7} \sec^{2} x d x

Limit will also change accordingly.
When

x \to 0

then

t \to 1

when

x \to \frac{π}{4}

then

t \to 8

Substituting the value,

\int_{0}^{\frac{π}{4}} \frac{\sec^{2} x}{{({(1 + 7 \tan x)}^{\frac{2}{3}})}^{\frac{2}{3}}} = \int_{1}^{8} \frac{1}{{(t)}^{\frac{1}{{(t)}^{\frac{2}{3}}}}} \cdot \frac{d t}{7}

= \frac{1}{7} \int_{1}^{8} {(t)}^{- \frac{2}{3}} d t

= \frac{1}{7} {[\frac{t^{- \frac{2}{3} + 1}}{- \frac{2}{3} + 1}]}_{1}^{8}

= \frac{1}{7} {[\frac{t^{\frac{1}{3}}}{\frac{1}{3}}]}_{1}^{8}

= \frac{3}{7} [8^{\frac{1}{3} - 1^{\frac{1}{3}}}}]

= \frac{3}{7} [2 - 1]

= \frac{3}{7}

Hence required answer is

\frac{3}{7}

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